OBJECTIVES

Having completed this chapter you should be able to:

■ Recognize a quadratic expression, understand how it arises and how to factorize it.

■ Solve a quadratic expression using trial-and-error methods and also by using the

formula.

■ Understand why some quadratic expressions cannot be factorized and why some

have a single repeated factor.

■ Recognize a quadratic function and sketch its graph.

■ Understand the relationship between the algebraic and graphical solution of a

quadratic equation.

■ Recognize and solve simultaneous quadratic equations.

■ Recognize and understand the role of quadratic functions in supply and demand

models and as total cost or total revenue functions.

Chapter 4

Quadratic equations

4.1 IntroductionIn the previous chapter we examined linear equations and functions. We saw that a linear equation has the general form ax + b = c, where x is the variable or unknown and a, b, and c areconstants or parameters. This equation is said to be linear because x is raised only to the power1, and to no higher or lower power. A linear function has the general form y = ax + b andspecifies a relationship between the two variables or unknowns, x and y. This function is linear,in this case because x and y are both raised only to the power 1, and to no higher or lower power.We also saw that the graph of the function y = ax + b is a straight line, which gives a concretemeaning to the term ‘linear’.

As we saw in the previous chapter, the key feature of a linear relationship between two vari-ables x and y is that the slope of the graph is constant, meaning that a 1-unit increase in x alwaysincreases or decreases y by the same amount. While some relationships in economics may be ofthis simple linear type, obviously we need to develop the ability to handle non-linear relation-ships too, where a given change in x does not always have the same effect on y. In this chapter

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we will examine quadratic equations and quadratic functions, which are the simplest type ofnon-linear relationship. We will show that a quadratic function, when graphed, produces acharacteristically U-shaped curve. We will show how to solve quadratic equations, including simultaneous quadratic equations. At the end of the chapter we look briefly at two applicationsof quadratic functions in economics.

In chapter 5 we will briefly examine some other non-linear functional relationships. Fortu-nately, it turns out that knowledge of only a handful of function types is sufficient to carry usthrough to quite an advanced level of economic analysis. So, although you may find this chapterand the next somewhat challenging, you can take comfort from the fact that what you will learnhere will, if understood and digested, carry you a long way in your study of economics.

4.2 Quadratic expressionsExpanding (multiplying out) pairs of brackets

Consider an expression such as

(a + b)(c + d)

where a, b, c, and d are unspecified constants.This is a product: that is, it is (c + d) multiplied by (a + b). We can think of this product as

measuring the area of a rectangle with width of (a + b) and height of (c + d), as in figure 4.1.But we see from figure 4.2 that the total area can be broken down into two sub-rectangles. The

rectangle on the left has width a and height c + d, so its area is a(c + d). The rectangle on the righthas width b and height c + d, so its area is b(c + d). Therefore the total area is a(c + d) + b(c + d).

Further, from figure 4.3 we can see that the total area can alsobe broken down into four sub-rectangles. The top left-handsub-rectangle has a width of a and a height of c, so its area is ac.Similarly, the other three sub-rectangles have areas of ad, bc, and bd.

Therefore the area of the rectangle is

(a + b)(c + d) (from figure 4.1)a(c + d) + b(c + d) (from figure 4.2)ac + ad + bc + bd (from figure 4.3)

Therefore we have:

(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd

This tells us how to ‘expand’ (multiply out) an expression suchas (a + b)(c + d). In step 1 we multiply (c + d) first by a, and thenby b, and add the results. This gives us figure 4.2: that is,

In step 2 we simply multiply out the brackets on the right-handside, in the way we explained in section 2.5. This gives us figure4.3: that is,

a(c + d) + b(c + d) = ac + ad + bc + bd

In words, each term in the second bracket (c and d) is multipliedby each term in the first bracket (a and b), and the results areadded.

(a + b)(c + d) = a(c + d) + b(c + d)

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a + b

c + darea = (a + b)(c + d)

Figure 4.1

a b

c + darea = a(c + d)area =

b(c + d)

Figure 4.2

a b

c

d

area = ac area = bc

area = bdarea = ad

Figure 4.3

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So the rule is:

RULE 4.1 Multiplying out brackets

(a + b)(c + d) ≡ a(c + d) + b(c + d)≡ ac + ad + bc + bd

Note that rule 4.1 holds for any values of a, b, c, and d. In other words, the relationships in rule4.1 are not equations, but identities. That is why we have used the ‘≡’ rather than the ‘=’ sign.(Refer back to section 3.1 if you are uncertain about this distinction.)

Hint When applying this rule, be careful to obey the sign rules (see rule 2.1 in chapter 2). For example:

(a − b)(c + d) = a(c + d) − b(c + d)= ac + ad − bc − bd

Note the change of sign when multiplying out −b(c + d)

A special case of rule 4.1 arises when a = c and b = d. Then we have (a + b)(a + b) ≡ (a + b)2. Thisis shown in rule 4.2:

RULE 4.2 Multiplying out brackets: a special case

(a + b)(a + b) ≡ (a + b)2

≡ a(a + b) + b(a + b)≡ a2 + ab + ba + b2

≡ a2 + 2ab + b2

As we can see in figure 4.4, rule 4.2 gives us the area of a squarewith sides of length a + b.

In our discussion up to this point, we have left open the ques-tion whether a and b are variables or constants, as this questionhas no effect on the reasoning that leads to rules 4.1 and 4.2.However, we shall frequently meet expressions of the form

(x + a)(x + b)

where x is a variable or unknown, and a and b are constants orparameters. (Refer back to section 3.3 if you are unsure of themeaning of any of these words.)

Applying rule 4.1 to this expression, we get rule 4.3:

RULE 4.3 Multiplying out brackets: another special case

(x + a)(x + b) ≡ x(x + b) + a(x + b)≡ x2 + xb + ax + ab≡ x2 + (a + b)x + ab

It is worth taking careful note of the pattern in rule 4.3, as we shall meet it frequently. This ruletells us that

(x + a)(x + b) ≡ x2 + (a + b)x + ab

The right-hand side, x2 + (a + b)x + ab, is an example of what is called a quadratic expression.

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QUADRATIC EQUATIONS

a b

a

b

area = a2 area = ab

area = b2 area = ab

Figure 4.4

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A quadratic expression by definition contains an unknown, x, raised to the power 2 (that is,x2). It may also contain a term in which x is raise to the power 1 (that is, x), but it must not contain any term in which x is raised to any higher or lower powers. It may also contain a constant, which is ab in the form above. Note especially the use of the ‘≡’ sign, which reminds us that the above equation is an identity : that is, it is true irrespective of the values of x, a, and b.(As a historical note, this came to be called a quadratic expression because it gives us the area ofa rectangle with one side of length x + a and the other side of length x + b. A rectangle has foursides; and ‘quad’ means four.)

Progress exercise 4.1

1. Explain why (a + b)2 does not equal a2 + b2.

2. Expand (= multiply out) the following:

(a) (x + 3)(x + 2) (b) (x + 5)(x − 4) (c) (2x + 1)(x + )

(d) (x − 4)2 (e) (x + 5)(x − 5)

4.3 Factorizing quadratic expressionsRule 4.3 tells us that if we expand (multiply out) the expression

(x + a)(x + b)

we get

x2 + (a + b)x + ab

which is called a quadratic expression.Factorizing a quadratic expression is simply the reverse of this process. We start with

x2 + (a + b)x + ab

and work back to find the factors of this expression, (x + a) and (x + b).

Hint If you are feeling a little lost, remember that 4 and 3 are factors of 12 because 4 × 3 ≡ 12.In the same way, from rule 4.3, (x + a) and (x + b) are the factors of x2 + (a + b)x + ab because

(x + a)(x + b) ≡ (x + a) × (x + b) ≡ x2 + (a + b)x + ab

Technique for factorizing a quadratic expression

EXAMPLE 4.1

Suppose we are asked to factorize

x2 + 9x + 20

Let us assume that the factors are (x + A) and (x + B), where A and B are two numbers thatwe haven’t yet found (but are hoping to find in the near future!).

Given this, then from the definition of factors we know that

(x + A)(x + B) = x2 + 9x + 20 (4.1)

But we also know from rule 4.3 that

(x + A)(x + B) = x2 + (A + B)x + AB (4.2)

(Check for yourself that equation (4.2) is correct.)

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Since the left-hand sides of equations (4.1) and (4.2) are identical, their right-hand sidesmust be identical too. Therefore

x2 + 9x + 20 = x2 + (A + B)x + AB

But this can only be true if A + B = 9 and AB = 20.

So we are looking for two numbers, A and B, whose sum is 9 and whose product is 20.After a little thought, it occurs to us that this will be true if A = 5 and B = 4, for then wehave A + B = 5 + 4 = 9, and AB = 5 × 4 = 20.

Therefore the factors are (x + 5) and (x + 4). We can check that this is correct by expanding

(x + 5)(x + 4) which by rule 4.3 gives

(x + 5)(x + 4) = x2 + 9x + 20

which is what we were asked to factorize.

EXAMPLE 4.2

x2 − 5x + 6

Following the same procedure as in example 4.1, we get

(x + A)(x + B) = x2 − 5x + 6 (4.3)

and

(x + A)(x + B) = x2 + (A + B)x + AB (4.4)

Comparing the right-hand sides of the two equations, we see that we must have A + B = −5and AB = 6. This example is a little trickier than example 4.1 because of the minus sign. Butwe can deduce that A and B must have the same sign (that is, both positive or bothnegative) because their product, AB, is positive.

We can list some possible values of A and B, given that they must have the same sign andthat their product, AB, must equal 6:

AB A B A ++ B

6 6 1 7

6 3 2 5

6 −6 −1 −7

6 −3 −2 −5

Clearly it is only the last of these that gives the correct values for both AB and A + B. So weconclude that A = −3 and B = −2. Therefore the factors are (x − 3) and (x − 2). We checkthis using rule 4.3:

(x − 3)(x − 2) = x2 − 2x − 3x + 6= x2 − 5x + 6

which is what we were asked to factorize.

EXAMPLE 4.3

3x2 − 6x − 72

In this case we first take the 3 outside brackets, giving us

3x2 − 6x − 72 = 3(x2 − 2x − 24)

On the right-hand side, we ignore the 3 for the moment and look for the factors of x2 − 2x − 24. Following the same procedure as in example 4.1, we get

(x + A)(x + B) = x2 − 2x − 24 (4.5)

and

(x + A)(x + B) = x2 + (A + B)x + AB (4.6)

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Comparing the right-hand sides of the two equations, we see that we must have A + B = −2and AB = −24 This case is a little complicated because both A + B and AB are negative. Butwe can deduce that A and B must have opposite signs (that is, one positive and the othernegative) because their product, AB, is negative. So let us assume that A is negative and Bpositive. Also, if A is negative and B positive, then A must be greater in absolute magnitudethan B because their sum, A + B, is negative.

Using these inferences, we can list some possible values of A and B, given that their productmust be −24:

AB A B A ++ B

−24 −24 1 −23

−24 −12 2 −10

−24 −8 3 −5

−24 −6 4 −2

We see that it is only the last of these that gives the right values for both AB and A + B. So we conclude that the factors of x2 − 2x − 24 are (x − 6) and (x + 4). That is,

x2 − 2x − 24 = (x − 6)(x + 4)

Multiplying both sides by 3, we get

3x2 − 6x − 72 = 3(x − 6)(x + 4)

Thus the factors of 3x2 − 6x − 72 are 3, (x − 6), and (x + 4).

Hint Not every quadratic expression can be factorized. Later we shall see exactly whatconditions must be fulfilled in order that a quadratic expression may be factorized.

Clearly, the method of factorizing quadratic expressions used above is rather unsatisfactorybecause it relies on essentially ‘trial and error’ methods. Fortunately, there is a much more powerful method of factorizing quadratic expressions, which we examine shortly.

Meanwhile, you may be wondering why we should want to factorize quadratic expressions.This will become clear in the next section.

4.4 Quadratic equations

EXAMPLE 4.4

In the previous section we considered quadratic expressions such as x2 − 2x − 24. Thesecontained no ‘=’ sign, and therefore were not equations. Suppose now we want to solve the equation

x2 + 9x + 20 = 0 (4.7)

This is an example of a quadratic equation. Equation (4.7) is quadratic because the left-handside is a quadratic expression, as defined at the end of section 4.2 above. It is quadratic becausex is raised only to the power 2 and the power 1 (because 9x ≡ 9x1), and to no other powers.

To solve it, we factorize the left-hand side. Using the ‘trial and error’ methods of theprevious section, we find that the factors are (x + 5) and (x + 4). So we know that

x2 + 9x + 20 = (x + 5)(x + 4)

Moreover, as noted earlier, this equation is an identity: that is, it is true for all values of x.This fact permits us to replace x2 + 9x + 20 in equation (4.7) with something identical,namely the product (x + 5)(x + 4). This gives

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(x + 5)(x + 4) = 0 (4.8)

The key point is that since equation (4.8) is identical to equation (4.7), it follows that anyvalue of x that satisfies (4.8) will automatically satisfy (4.7). In other words, any value of xthat is a solution to (4.8) is also a solution to (4.7). So if we can solve (4.8), we have alsosolved (4.7).

So now we want to solve equation (4.8). Why is (4.8) any easier to solve than (4.7)? Theanswer is: because the left-hand side of (4.8) is a product. We have something of the form

M × N = 0

where M = (x + 5) and N = (x + 4).

Now, the product M × N equals zero whenever either M = 0 or N = 0 (because ‘zero timesanything is zero’—see section 1.3 if you are uncertain on this point).

Thus the product (x + 5)(x + 4) equals zero whenever either (x + 5) = 0 or (x + 4) = 0. So

(x + 5)(x + 4) = 0 (4.8) repeated

is satisfied when

either (x + 5) = 0 (which is true when x = −5)

or (x + 4) = 0 (which is true when x = −4)

Therefore we conclude that x = −5 and x = −4 are the solutions (also known as roots) toequation (4.7).

We can check that these solutions are correct by substituting each of them in turn intoequation (4.7). When x = −5, equation (4.7) becomes the identity

(−5)2 + 9(−5) + 20 = 25 − 45 + 20 = 0

and when x = −4, equation (4.7) again becomes the identity

(−4)2 + 9(−4) + 20 = 16 − 36 + 20 = 0

To summarize this example, to solve the quadratic equation x2 + 9x + 20 = 0, we firstfactorized the left-hand side and obtained (x + 5)(x + 4) = 0. This equation is satisfiedwhen x = −5 or x = −4. These values are therefore the solutions (or roots, as they areusually called) to the quadratic equation x2 + 9x + 20 = 0.

Let’s look at another example of solving a quadratic equation.

EXAMPLE 4.5

3x2 − 6x − 72 = 0 (4.9)

To solve equation (4.9), we factorize the left-hand side. From example 4.3, we know thatthe factors of 3x2 − 6x − 72 = 0 are 3, (x − 6) and (x + 4). So we know that

3x2 − 6x − 72 ≡ 3(x − 6)(x + 4)

and moreover this equation is an identity—it is true for all values of x. Therefore, whenever3(x − 6)(x + 4) equals zero, 3x2 − 6x − 72 also equals zero because the two are identicallyequal.

Now, 3(x − 6)(x + 4) = 0 when either x = 6 or x = −4. (Check this for yourself.) Therefore we conclude that x = 6 and x = −4 are solutions to equation (4.9).

We can check that these solutions are correct by substituting each of them in turn intoequation (4.9). When x = 6, equation (4.9) becomes the identity

3(6)2 − 6(6) − 72 = 108 − 36 − 72 = 0

When x = −4, equation (4.9) again becomes the identity:

3(−4)2 − 6(−4) − 72 = 48 + 24 − 72 = 0

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Generalization

To solve any quadratic equation of the form

ax2 + bx + c = 0 (where a, b, and c are given constants)

we first take the a outside brackets, giving us

Next, we look for the factors of x2 + x + . Let us assume that we find the factors to be

(x + A) and (x + B) (where A and B are two numbers)

If these are the factors, then by definition we have

After multiplying both sides by a, this becomes

a(x + A)(x + B) ≡ ax2 + bx + c

As this is an identity, when the left-hand side equals zero, the right-hand side must equal zerotoo. And the left-hand side equals zero when x = −A or x = −B. These values are therefore the solutions to the quadratic equation ax2 + bx + c = 0.

4.5 The formula for solving any quadratic equationIn the previous two sections we factorized quadratic expressions by ‘trial and error’ methods.This is not very satisfactory as it relies too much on guesswork and inspiration. Fortunately, a formula exists (which we will not prove here) for solving any quadratic equation. This takesus straight to the solutions (also known as roots), and also tells us whether a solution exists. Theformula is as follows:

RULE 4.4 Solving a quadratic equation

Given any quadratic equation

ax2 + bx + c = 0 (where a, b, and c are given constants)

the solutions (roots) are given by the formula

Notice in the formula the symbol ‘±’, meaning ‘plus or minus’. Thus the formula generates thenecessary two solutions, as we take first the ‘plus’ option, then the ‘minus’. This will becomeclearer in the examples below.

This formula looks rather daunting at first sight, but if you say it aloud a few times you willsoon find you have memorized it. And with a little practice you should have no difficulty in applying it correctly. Here are two examples of rule 4.4 in action:

EXAMPLE 4.6

x2 + 5x + 6 = 0

Comparing this with the standard form

ax2 + bx + c = 0

x

b b ac

a=

±– –2 4

2

( )( )x A x B x

b

ax

c

a+ + ≡ + +2

ca

ba

a xb

ax

c

a2 0+ +

⎛

⎝⎜

⎞

⎠⎟ =

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we see that a = 1, b = 5, and c = 6. Substituting these values into the formula (see rule 4.4),we get

(note that the square root of 1 is 1)

= either −2 or −3

These solutions (roots) can be checked by substituting each in turn into the givenequation. Be sure to check this for yourself.

Notice also that these solutions imply that the factors of x2 + 5x + 6 are (x + 2) and (x + 3).We can confirm this by expanding (x + 2)(x + 3), which gives

(x + 2)(x + 3) ≡ x2 + 3x + 2x + 6 ≡ x2 + 5x + 6 (as we expected).

Note we have used the ‘≡’ sign here to indicate an identity (see section 4.2 and especiallyrule 4.1 above).

EXAMPLE 4.7

x2 + 4x − 12 = 0

Comparing this with the standard form

ax2 + bx + c = 0

we see that a = 1, b = 4, and c = −12. Substituting these values into the formula in rule 4.4,we get

= either 2 or −6

These solutions (roots) can be checked by substituting each in turn into the givenequation. Be sure to check this for yourself.

The solutions imply that the factors of x2 + 4x − 12 are x − 2 and x + 6. We can confirm thisby expanding (x − 2)(x + 6), which gives

(x − 2)(x + 6) ≡ x2 + 6x − 2x − 12 ≡ x2 + 4x − 12

Note once more that the ‘≡’ sign indicates an identity (true for all values of x).

4.6 Cases where a quadratic expression cannot be factorizedIn section 4.3 above we said that not every quadratic expression could be factorized. Now weare ready to see why. Consider the following example:

EXAMPLE 4.8

x2 + 2x + 2 = 0

Let us see what happens when we try to solve this equation using the formula. In this case wehave a = 1, b = 2, and c = 2. Substituting these values into the formula (see rule 4.4) gives

=

±=

+– – – –4 64

2

4 8

2

4 8

2either or (note that 664 8= )

xb b ac

a=

±=

±=

± +– – – – ( )(– )

( )

–2 24

2

4 4 4 1 12

2 1

4 16 48

2

=

±=

+–

– – –5 1

2

5 1

2

5 1

2either or

xb b ac

a=

±=

±=

±– – – – ( )( )

( )

– –2 24

2

5 5 4 1 6

2 1

5 25 24

2

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At this point we grind to a halt. This is because, in the numerator, we have , so to goany further requires us to find the square root of minus 4, and we know from section 1.14that a negative number has no square root. Thus we conclude that the equation x2 + 2x + 2 = 0 has no solutions (roots).

Actually the conclusion of example 4.8 is not quite correct, as it stands. Mathematicians, whohave very tidy minds, found it very unsatisfactory that some quadratic equations had solutionsand some did not. They therefore defined a new type of number, called an imaginary number,which has the property that its square is negative. (We mentioned this in section 1.14.) Wewon’t go into the details here, but this definition makes it possible to conclude that the equa-tion above does, in fact, have roots, but these roots are not real numbers. Thus the formally cor-rect conclusion is that the equation x2 + 2x + 2 = 0 has no real roots. We examine this questionmore closely in chapter W21, which is available at the Online Resource Centre.

Generalization

The case above arose because, in the formula (rule 4.4), the expression under the square rootsign was negative. When we look again at the formula

it is clear that this problem will arise whenever b2 is less than 4ac, for then b2 − 4ac, the expres-sion under the square root sign, is negative. For example, when a = 1, b = 4, c = 5, then b2 = 16and 4ac = 20, so b2 − 4ac = 16 − 20 = −4.

So we conclude that the quadratic equation ax2 + bx + c = 0 has no real solutions (roots) ifthe values of the parameters a, b, and c are such as to make b2 less than 4ac.

4.7 The case of the perfect square

EXAMPLE 4.9

x2 + 6x + 9 = 0

Let’s see what happens when we try to solve this, using the formula. In this case we have a = 1, b = 6, and c = 9. Substituting these values into the formula gives

Thus the equation has only one solution (root), −3. This is because the expression under the square root sign happens, in this case, to equal zero (and the square root of zero is zero). It therefore disappears (‘drops out’) from the solution, leaving only one solution.

=

± −= =

– ––

6 36 36

2

6

23

xb b ac

a=

±=

±– – – – ( )( )

( )

2 24

2

6 6 4 1 9

2 1

x

b b ac

a=

±– –2 4

2

−4

=

±=

±– – – –2 4 8

2

2 4

2

xb b ac

a=

±=

±– – – – ( )( )

( )

2 24

2

2 2 4 1 2

2 1

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You might think this means that x2 + 6x + 9 has only one factor, (x + 3). But this can’t beright, since we must multiply (x + 3) by something in order to produce x2 + 6x + 9. Thatsomething is (x + 3). In other words, we have

(x + 3)(x + 3) = x2 + 6x + 9

The expression x2 + 6x + 9 is called a perfect square because it equals the area of a squarewith sides of length (x + 3). The equation x2 + 6x + 9x = 0 has two equal solutions, alsodescribed as a single, repeated solution.

Generalization

The case above arose because, in the formula, the expression under the square root sign waszero. When we look again at the formula

it is clear that this problem will arise whenever the parameters a, b, and c of the equation ax2 +bx + c = 0 are such as to make b2 equal to 4ac. The formula then collapses down to

For example, when a = 2, b = 4, c = 2, then b2 = 4ac = 16, so b2 − 4ac = 16 − 16 = 0.

Summary of sections 4.1–4.7

When (x + A)(x + B) is expanded (multiplied out) it becomes the quadratic expression x2 +(A + B)x + AB (see rule 4.3). Then (x + A) and (x + B) are the factors of x2 + (A + B)x + AB.

We factorize x2 + (A + B)x + AB by finding two numbers, A and B, such that A + B is thecoefficient of x and AB is the constant term. The solutions (roots) to the quadratic equation x2 + (A + B)x + AB = 0 are therefore x = −A and x = −B.

However, the formula in rule 4.4 gives us an easier method of solving quadratic equations ofthe form ax2 + bx + c = 0. There are three classes of solution to this equation, depending on thevalues of the parameters, a, b, and c:

(1) If b2 > 4ac, there are two distinct roots.

(2) If b2 = 4ac, there is only one root, described as a repeated root.

(3) b2 < 4ac, there are no roots (more precisely, there are no roots that are real numbers, butroots may be found by defining a new type of number, an imaginary number).

Progress exercise 4.2

1. By trial and error, solve the following quadratic equations (where possible). Then checkyour answers by solving them using the formula.

(a) x2 − 3x + 2 = 0 (b) x2 + 5x + 4 = 0 (c) x2+ 4x − 5 = 0

(d) x2 + x + = 0 (e) x2 − 16 = 0

(f) 2x2 + 8x + 6 = 0 Hint: if ax2 + bx + c = 0, then x2 + x + = 0

(g) −x2 − 0.5x + 3 = 0 (h) x2 + 8x + 16 = 0 (i) x2 − x − 2 = 0 (j) x2 + 3x + 3 = 0

What makes the answers to (h) and (j) different from the others?

ca

ba

18

34

x

b

a=

–

2

x

b b ac

a=

±– –2 4

2

1194

QUADRATIC EQUATIONS

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4.8 Quadratic functionsIf we write, for example

y = x2 + 5x + 6

we have a function: that is, a relationship between an independent variable, x, and a dependentvariable, y. (See section 3.5 if you are unsure what a function is.) Because x is raised to the power2 (and to no higher power), this is called a quadratic function. The general form of thequadratic function is

y = ax2 + bx + c (where a, b, and c are parameters)

The example above, y = x2 + 5x + 6, is obtained from the general form by setting a = 1, b = 5,and c = 6.

Let’s consider what the graph of a quadratic function will look like, by looking at some examples.

EXAMPLE 4.10

Starting from the general form, y = ax2 + bx + c, let’s take the case where a = 1, b = 0 and c = 0. This gives us the simplest case of a quadratic function:

y = x2

If we draw up a table of values (table 4.1) for values of x between −4 and +4, and plot theresulting values as a graph, the result is seen in figure 4.5. Notice that y is positive when x iseither positive or negative. When x is zero, y is also zero.

This curve is called a parabola. The U-shape arises from the fact that x2 (and therefore y) is positive when x is either positive or negative. For example, y = 4 when x = +2 and alsowhen x = −2. In general, y has the same value when x = x0 as it does when x = −x0.Therefore the curve has two arms or branches, lying symmetrically about the y-axis. If you fold the graph paper along the y-axis, the two branches of the graph meet.

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Table 4.1 Values for graph of y = x2.

x −4 −3 −2 −1 0 1 2 3 4

y = x2 (−4)2 (−3)2 (−2)2 (−1)2 (0)2 (1)2 (2)2 (3)2 (4)2

= 16 = 9 = 4 = 1 = 0 = 1 = 4 = 9 = 16

EXAMPLE 4.11

If we repeat example 4.10, but this time with a = 3 instead of a = 1, we obtain

y = 3x2

Table 4.2 gives values of y for x between −4 and +4, and the resulting curve is also plottedin figure 4.5. From the table, and also in the graph, we can see that, for any given value of x,y is now three times as large as in example 4.10. For example, when x = +2 or −2, the curvey = x2 gives y = 4, while the curve y = 3x2 gives y = 3 × 4 = 12.

Table 4.2 Values for graph of y = 3x2.

x −4 −3 −2 −1 0 1 2 3 4

y = 3x2 3(−4)2 3(−3)2 3(−2)2 3(−1)2 3(0)2 3(1)2 3(2)2 3(3)2 3(4)2

= 48 = 27 = 12 = 3 = 0 = 3 = 12 = 27 = 48

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EXAMPLE 4.12

If we repeat example 4.10 but with a = −1 instead of a = +1, we get

y = −x2

The effect is that, compared with example 4.10, the sign of y associated with any givenvalue of x is reversed. For example, when x = −3, the function y = x2 gives y = (−3)2 = +9,while the function y = −x2 gives y = −[(−3)2] = −9. This inverts the graph (see figure 4.6).We haven’t bothered to compile a new table of values, as it would be exactly the same asthe table for y = x2, but with every value of y negative instead of positive. At x = 0 the twocurves coincide, of course.

EXAMPLE 4.13

This repeats example 4.10, but this time we set c = 4. This gives us

y = x2 + 4

Compared with example 4.10, we have added a constant, 4, so it should be fairly obviousthat this shifts the whole curve bodily upwards by 4 units (see figure 4.7). The effect of thisis that the curve now cuts (intercepts) the y-axis at y = 4. Similarly, if we had set c = −4instead of +4, this would have shifted the whole curve bodily downwards by 4 units. Thecurve would then intercept the y-axis at y = −4 (see also figure 4.7).

EXAMPLE 4.14

If we repeat example 4.10 but with b = 3, we have

y = x2 + 3x

Let’s consider where this curve will lie in comparison with y = x2. When x is positive, soalso is 3x, and therefore the graph of y = x2 + 3x will lie above the graph of y = x2. Forexample, when x = 1, 3x = 3, and therefore when x = 1, the graph of y = x2 + 3x will lie 3units above the graph of y = x2 (see figure 4.8). Conversely, when x is negative, so also is 3x,

1214

QUADRATIC EQUATIONS

x1 2 3 4–4 –3 –2 –1 0

y = x2

y = 3x2

y

8

6

4

2

16

10

14

12

x1 2 3 4–4 –3 –2 –1 0

y = –x2

y

–8

–6

–4

–2

–16

–10

–14

–12

2

4

Figure 4.5 Graphs of y = x2 and y = 3x2. Figure 4.6 Graph of y = −x2.

9780199602124_109_133_CH04.qxd 10/7/11 8:53 Page 121

so the graph of y = x2 + 3x will lie below the graph of y = x2. When x = 0, 3x also equalszero, so y = 0 and the two curves coincide.

Thus the addition of the 3x term has the effect of shifting the curve up when x is positiveand down when x is negative. If we subtract 3x instead of adding it, the direction of shift isreversed: that is, the graph of y = x2 − 3x lies below the graph of y = x2 when x is positiveand above it when x is negative.

Summary

Reviewing the five previous examples, we can draw the following conclusions about the shapeof the quadratic function y = ax2 + bx + c.

(1) The x2 term gives the graph an approximate U-shape, called a parabola. If the parameter ais positive, the graph looks like a U. If the parameter a is negative, the graph looks like aninverted U (that is, like this: ∩). The absolute magnitude of a determines how steeply thecurve slopes up (or down).

(2) The constant term, c, determines the intercept of the curve on the y-axis.

(3) The x term shifts the parabola up and down. If the parameter b is positive, the curve shiftsup when x > 0 and down when x < 0. If b is negative, this shift is reversed. The absolute mag-nitude of b determines the strength of the shift effect.

4.9 The inverse quadratic functionHere we consider a question which may appear to have little point now, but which is importantlater in this book. It concerns a function that is a close relative of the quadratic function y = x2:

y = (which we can also write as y = x1/2 or y = x0.5)x

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x0

y = x2 + 4

y = x2 – 4

y

8

6

2

16

10

14

12

–2

1 2 3 4–4 –3 –1 x0

y = x2

y

1 2 3–4 –3 –2 –1–5

8

6

4

2

16

10

14

12

–2

–4

y = x2 + 3x

4

–4

–2

Figure 4.7 Graphs of y = x2 + 4 and y = x2 − 4. Figure 4.8 Graphs of y = x2 and y = x2 + 3x.

9780199602124_109_133_CH04.qxd 10/7/11 8:53 Page 122

What does the graph of this function look like? Toanswer this, we first consider the function x = y2.This is a quadratic function like y = x2, except thatit has x instead of y as the dependent variable. Soits graph is the usual U-shaped curve but nowwith its two branches or arms oriented towardsthe x-axis rather than the y-axis, as in figure 4.9.

If we now take square roots on both sides of x = y2, it becomes y = . Because the square rootof any number is positive, by definition (see sections 1.14 and 2.9), y is positive. The graph of y = is shown in figure 4.9. It is that part of the graph of x = y2 lying above the x-axis; that is,where y is positive. This is an important function,for it gives us the relationship between any posi-tive number, x, and its square root, y. The part ofthe graph of x = y2 lying below the x-axis in figure4.9 is the graph of another function, y = − .

Thus the function x = y2 does not have a trueinverse. The branch of the function x = y2 wherey is positive has y = as its inverse; and the branchwhere y is negative has y = − as its inverse.

You might be thinking that we could splice together the graphs of y = and y = − to

make a combined graph, y = ± . This would include both positive and negative values of y;for example, when x = 4, we would have y = ±2 (see dotted lines in figure 4.9). Thus perhaps y = ± could be considered to be the inverse function to x = y2.

The problem with this is that y = ± is not a function, but merely a relation, a somewhatlooser connection between two variables. This is because the definition of a function requiresthat there be no more than one value of the dependent variable (y in this case) associated withany given value of x. This condition is not satisfied in the case of y = ± . For example, as we have just seen, when x = 4, y may be either +2 or −2. Thus there are two values of y associ-ated with any positive value of x. So we must continue to view y = and y = − as distinctfunctions.

These ideas extend straightforwardly to other even-numbered roots, such as y = , but wewill not pursue this here. The problem of a non-existent inverse function does not arise withodd numbered roots. For example, as we shall see in the next chapter, the inverse of the func-tion x = y3 is the function y = .

4.10 Graphical solution of quadratic equations

EXAMPLE 4.15

Let us consider the function y = x2 + x − 6, which is graphed in figure 4.10.

We see that the graph cuts the x-axis at x = +2 and x = −3. Recall that, everywhere along thex-axis, y = 0 (see section 3.6 if you are uncertain on this point). Therefore y = 0 at these twopoints. But we know also that y = x2 + x − 6 at these two points, because they lie on thecurve. Thus two things are true at the two points, x = +2 and x = −3:

y = 0 and y = x2 + x − 6

x3

x4

xx

x

xx

xxx

xx

x

x

x

1234

QUADRATIC EQUATIONS

x

y = –√x

y = √x or x = y2

y

4

3

2

1

–1

–2

–3

–4

0 2 4 6 8 10 12 14 16

Figure 4.9 Graphs of y = and y = − .xx

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Combining these two facts, we canconclude that x2 + x − 6 = 0 at thesepoints. Therefore x = +2 and −3 arethe solutions to the quadraticequation x2 + x + 6 = 0.

Thus the values of x at which thegraph of the quadratic functiony = x2 + x − 6 cuts the x-axis give us the solutions (roots) of thequadratic equation x2 + x − 6 = 0.

This exactly parallels the case of the linear function that weexamined in section 3.8, except that we now have two solutionsbecause the graph cuts the x-axistwice.

We can check this result algebraically. The equation x2 + x − 6 = 0 factorizes easily as

(x − 2)(x + 3) = 0

which is true when either x = +2 or x = −3. Thus the algebra confirms the information weobtained from the graph.

EXAMPLE 4.16

Consider the function y = −2x2 + 3x + 5, which is graphed in figure 4.11.

The graph cuts the x-axis at x = −1 and x = 2 . Therefore y = 0 at these two points, becausethey lie on the x-axis. But we know also that y = −2x2 + 3x + 5 at these two points, becausethey lie on the curve. Combining these two facts, we can conclude that −2x2 + 3x + 5 = 0 atthese points.

Therefore x = −1 and x = 2 are the solutions to the quadratic equation −2x2 + 3x + 5 = 0.

Thus, as in example 4.15, the values of x at which the graph of the quadratic function y = −2x2 + 3x + 5 cuts the x-axis give us the solutions (roots) of the quadratic equation−2x2 + 3x + 5 = 0.

We can check this result algebraically. Using rule 4.4 with a = −2, b = 3, and c = 5, thesolution to the equation −2x2 + 3x + 5 = 0 is

Thus the algebra and the graph corroborate one another.

EXAMPLE 4.17

The function y = x2 + 6x + 12 is graphed in figure 4.12. As the graph does not cut the x-axisat any point, we can conclude immediately that the equation x2 + 6x + 12 = 0 has no (real)solutions. This is confirmed when we apply rule 4.4 with a = 1, b = 6, and c = 12, for wethen get

Since we have a negative number under the square root sign, this equation has no solutionsthat are real numbers.

x

b b ac

a=

±=

±=

±– – – – ( )( ) – –2 4

2

6 36 4 1 12

2

6 12

2

xb b ac

a=

±=

±=

±= −

– – – – (– )( )

–

–

–

2 4

2

3 9 4 2 5

4

3 49

41 or 2 1

2

12

12

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Sy

y = x2 + x − 6

–10

–5

5

10

15

y = 0, andy = x 2 + x − 6Thereforex 2 + x − 6 = 0

y = 0, andy = x 2 + x − 6Thereforex 2 + x − 6 = 0

x–5 –4 –3 –2 –1 0 1 2 3 4 5

√

√

Figure 4.10 Graph of y = x2 + x − 6.

9780199602124_109_133_CH04.qxd 10/7/11 8:53 Page 124

We can conclude from example 4.17 that, in general, when the graph of a quadratic functionlies entirely above or below the x-axis, the corresponding quadratic equation has no real solu-tions. As mentioned in section 4.6, solutions for such equations can be found by defining animaginary number which has the property that its square is negative. For further discussion, seechapter W21 section 4, to be found at the Online Resource Centre.

EXAMPLE 4.18

The function y = x2 + 6x + 9 is also graphed in figure 4.12. We see that the graph is tangentto the x-axis at x = −3. This is the case of the ‘perfect square’ discussed in section 4.7 above.The corresponding quadratic equation is

x2 + 6x + 9 = 0

Applying rule 4.4 with a = 1, b = 6, and c = 9, we get

Because b2 = 4ac, the formula collapses to a single solution, −3, which we call a repeatedsolution.

We can also arrive at this conclusion by factorizing x2 + 6x + 9 = 0. It factorizes as (x + 3)(x + 3) = 0, so we have a single (repeated) solution, x = −3.

It is easy to see that example 4.18 generalizes. Whenever the graph of a quadratic function y = ax2 + bx + c is tangent to the x-axis at a point, we know that the corresponding quadraticequation, ax2 + bx + c = 0, is a perfect square.

We saw in section 4.7 that the quadratic expression ax2 + bx + c is a perfect square if b2 = 4ac.The single repeated solution is then x = − . In example 4.18 we have b2 = 4ac = 36.

At the Online Resource Centre we show how, using Excel®, quadratic functions can be graphedand quadratic equations solved. www.oxfordtextbooks.co.uk/orc/renshaw3e/

b2a

x

b b ac

a=

±=

±= =

– – – – ( )( ) ––

2 4

2

6 36 4 1 9

2

6

23

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QUADRATIC EQUATIONS

y = 0, andy = −2x2 + 3x + 5Therefore−2x2 + 3x + 5 = 0

y

x

y = −2x2 + 3x + 5

–25

–20

–15

–10

–5

5

10

–3 –2 –1 0 1 2 3 4

y = 0, and y = −2x2 + 3x + 5Therefore−2x2 + 3x + 5 = 0

y = x2 + 6x + 12

y = x2 + 6x + 9 2

4

6

8

10

12

16

14

18

20

–6 –5 –4 –3 –2 –1 0 1

x

y

The graph of y = x2 + 6x + 9 is tangent to the x-axis at x = −3.Therefore the quadratic equation x2 + 6x + 9 = 0 has a single solution, x = −3. Therefore factors of x2 + 6x + 9 are (x + 3) and (x + 3). That is, x2 + 6x + 9 ≡ (x + 3)(x + 3) ≡ (x + 3)2.

Figure 4.11 Graph of y = −2x2 + 3x + 5. Figure 4.12 Graphs of y = x2 + 6x + 12 and y = x2 + 6x + 9.

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Progress exercise 4.3

1. Sketch the graphs of the following quadratic functions. (Hint: it may help if you factorizethem.) Take values of x between −5 and +5 unless otherwise stated.

(a) y = x2 + x − 6 (b) y = −2x2 + 3x + 5 (c) y = x2 − 2x − 3

(d) y = 4x2 + 8x + 4 between x = −8 and x = 2 (e) y = x2 + 8

2. Use your graphs from question 1 to find approximate solutions to the following quadraticequations, where possible. Then check your answers by factorization or by applying the formula.

(a) x2 + x − 6 = 0 (b) −2x2 + 3x + 5 = 0 (c) x2 − 2x − 3 = 0

(d) 4x2 + 8x + 4 = 0 (e) x2 + 8 = 0

What makes (d) and (e) different from the others?

3. You are given the following information about a quadratic function y = ax2 + bx + c. In eachcase, find a, b, and c and sketch the graph.

(a) y = 0 when x = 2 or −3; y = −12 when x = 0. (b) a = −2; y = 0 when x = −4 or −2.

4. A quadratic function y = ax2 + bx + c has the following properties: a = −2, y = 0 when x = −3or −2. Find b and c and sketch the graph.

5. A quadratic function y = ax2 + bx + c has the following properties: a = −3; y = 0 when x = −2or x0; and y = 6 when x = 0. Find b, c, and x0 and sketch the graph.

4.11 Simultaneous quadratic equations

EXAMPLE 4.19

Suppose we are asked to solve the simultaneous equations

y = 2x2 + 3x + 2 (4.10)

y = x2 + 2x + 8 (4.11)

At first sight this appears rather difficult, because quadratic equations are still somewhatunfamiliar to us, and here we have two of the beasts! However, there is no reason why wecan’t solve this pair of simultaneous equations using exactly the same techniques that weused to solve simultaneous linear equations in section 3.9.

As we did in section 3.9, we can make use of the fact that, if the two equations are bothsatisfied simultaneously by a pair of values of x and y, the value of y in equation (4.10) willbe the same as the value of y in equation (4.11). Therefore, the left-hand side of equation(4.10) and the left-hand side of equation (4.11) will be equal to one another. But, if the twoleft-hand sides are equal, then the two right-hand sides must be equal to one another too.So we can simply set the right-hand sides of the two equations equal to one another, thensolve the resulting equation. This will give us the value of x that satisfies both equationssimultaneously. Doing this, we get

2x2 + 3x + 2 = x2 + 2x + 8

After subtracting the right-hand side from both sides, this becomes

2x2 + 3x + 2 − [x2 + 2x + 8] = 0 which simplifies to

x2 + x − 6 = 0

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We can solve this quadratic equation either by using ‘trial and error’ methods to findfactors, or by employing the formula (see rule 4.4 above). In this case, trial and errormethods quickly factorize the equation as

x2 + x − 6 ≡ (x + 3)(x − 2) = 0

so the solutions are x = −3 and x = +2. We can find the solution values for y by substitutingthese values into either equation (4.10) or equation (4.11). Using equation (4.10) we get

when x = −3, y = 2(−3)2 + 3(−3) + 2 = 11 and

when x = +2, y = 2(2)2 + 3(2) + 2 = 16

As a final check we should substitute x = −3, y = 11 and x = 2, y = 16 into both equations. Ifboth equations then become identities, our solutions are correct. (Check this for yourself.)

4.12 Graphical solution of simultaneous quadratic equationsIn the previous example we used algebraic methods to solve the simultaneous equations

y = 2x2 + 3x + 2

y = x2 + 2x + 8

and found the solutions to be x = −3, y = 11 and x = 2, y = 16.Now let us look at the graph of the two equations (see figure 4.13). From the graph, we see

that the two curves cut at x = −3, y = 11 and x = 2, y = 16. This is because any point where thetwo curves cut one another lies on both curves, and therefore the coordinates of that point mustsatisfy both equations simultaneously. This parallels exactly the reasoning of sections 3.9–3.10,where we examined linear simultaneous equations.

In general, therefore, the solutions to a pair of simultaneous quadratic equations are given bythe coordinates of the points of intersection of the two curves.

Simultaneous quadratic equations with no (real) solutions

We have just seen that the solutions to a pair of simultaneous quadratic equations are found atthe points where their graphs cut. If their graphs do not cut, the pair of simultaneous quadraticequations do not have a real solution. For example, consider the simultaneous quadratic equations

y = −2x2 − 3x − 4

y = x2 + 2x + 8

These two functions are graphed in figure 4.14, from which we see that they never intersect. So,as a pair of simultaneous quadratic equations they have no real solution. The two equations areinconsistent, as previously discussed for linear simultaneous equations (see section 3.11). We canconfirm this algebraically by setting the two right-hand sides equal to one another. This gives

−2x2 − 3x − 4 = x2 + 2x + 8 which rearranges as

3x2 + 5x + 12 = 0

Using rule 4.4, we get

The negative number under the square root sign tells us that there are no real solutions.

x

b b ac

a=

±=

±=

±– – – – ( )( ) – –2 4

2

5 25 4 3 12

6

5 119

6

1274

QUADRATIC EQUATIONS

9780199602124_109_133_CH04.qxd 10/7/11 8:53 Page 127

The Online Resource Centre shows how simultaneous quadratic equations can be graphed andsolved graphically using Excel®. www.oxfordtextbooks.co.uk/orc/renshaw3e/

Summary of sections 4.8–4.12

The graph of the quadratic function y = ax2 + bx + c has a characteristic approximate U-shape(if a > 0) or inverted U-shape (if a < 0). The solutions to the quadratic equation ax2 + bx + c = 0and found where the graph of y = ax2 + bx + c cuts the x-axis, for at these points y = 0. There arethree possibilities:

(1) If the graph of the function y = ax2 + bx + c cuts the x-axis at x = m and x = n, then these arethe two solutions (roots) of the corresponding quadratic equation ax2 + bx + c = 0.Therefore ax2 + bx + c = 0 ≡ (x − m)(x − n).

(2) If the graph of the function y = ax2 + bx + c is tangent to the x-axis at x = m, then this is thesingle (repeated) root solution of the corresponding quadratic equation ax2 + bx + c = 0.Therefore ax2 + bx + c = 0 ≡ (x − m)(x − m). The quadratic expression ax2 + bx + c is thensaid to be a perfect square.

(3) If the graph of the function y = ax2 + bx + c neither cuts nor is tangent to the x-axis, thenthe corresponding quadratic equation ax2 + bx + c = 0 has no (real) roots.

As with the linear functions in chapter 3, the graphical solution to a pair of simultaneousquadratic equations is found where the graphs of the two corresponding functions intersect.There may be 2, 1 or 0 (real) solutions.

4.13 Economic application 1: supply and demandAt the beginning of this chapter we said that we needed to develop some non-linear functionsin order to be able to analyse non-linear relationships in economics. We are not yet fullyequipped for this analysis, as we first need to look at some other non-linear functions in thenext chapter. However, to conclude this chapter on quadratic equations and functions we willbriefly indicate some economic relationships that might be quadratic in their form.

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y = –2x2 − 3x − 4

y = x2 + 2x + 8

–40

–30

–20

–10

10

20

30

–4 –3 –2 –1 1 2 3x

y

0

As their graphs never intersect, the simultaneousequations y = x2 + 2x + 8 and y = –2x2 − 3x − 4do not have a (real) solution. The equations areinconsistent.

y = 2x2 + 3x + 2

x

y = x2 +

2x

+ 8

(–3, 11)

(2, 16)

8

6

4

2

16

10

14

12

22

20

18

26

24

2 3–4 –3 –2 –1 0 1

Figure 4.13 Solution to y = 2x2 + 3x + 2; y = x2 + 2x + 8. Figure 4.14 Graphs of y = −2x2 − 3x − 4 and y = x2 + 2x + 8.

9780199602124_109_133_CH04.qxd 10/7/11 8:53 Page 128

We developed the idea of supply and demand functions in the previous chapter, includingfinding the equilibrium price and quantity and carrying out comparative static exercises, suchas finding the effects of a tax on the market equilibrium. There we were restricted to linear sup-ply and demand functions, but this was obviously very limiting. We can now extend the ana-lysis to cases where one or both of the supply and demand functions is quadratic in its form. We will work with the inverse demand and supply functions, as this is common usage amongeconomists (see section 3.15 if you need to revise this point).

EXAMPLE 4.20

A possible inverse demand function which is quadratic in form is

pD = 1.5q2 − 15q + 35

where pD is the demand price (see section 3.15). Let us also introduce an inverse supplyfunction, which for simplicity we’ll assume is linear:

pS = 2q + 7

where pS denotes the supply price. Note that, unlike in section 3.15, we will not distinguish between quantity demanded (qD) and quantity supplied (qS). Instead, we will simply take it as given that when we are looking at the demand function, qdenotes qD; and when at the supply function, that q denotes qS. We assume that inequilibrium qD = qS. The motive for this slight change in the model, compared to section 3.15, is simply to avoid the superscripts clashing with the powers, such as q2 in the demand function above.

The inverse demand and supply functions are graphed in figure 4.15. We see that thequadratic demand function has a fairly gentle curvature, becoming flatter as the price falls.This shape is plausible, since it implies a surge in demand as the price becomes very lowand the product appears a great bargain. However, we can see straight away that this formof quadratic function is not ideally suited to modelling demand. Quite reasonably, demandseems to reach its maximum at around q = 3.75 when p falls to zero, indicating that this isthe maximum quantity consumers wish to consume even when the product is free.However, the demand function then mysteriously pops up above the q-axis again ataround q = 6.25 and from then on is positively sloped, which is very hard to reconcile withany theory of consumer behaviour. To avoid this embarrassment we must restrict q tovalues less than, say, q = 4.

We can find the equilibrium price and quantity in this model in the same way as in section 3.15. Our model consists of the demand and supply functions above, together with the equilibrium condition, pD = pS. This simply says that the demand price (the price buyers are willing to pay for any given quantity) must equal the supplyprice (the price sellers are willing to accept for any given quantity). Given pD = pS, we can set the right-hand sides of the demand and supply functions equal to one another, giving

1.5q2 − 15q + 35 = 2q + 7

This rearranges to

1.5q2 − 17q + 28 = 0

Solving this quadratic using the formula (rule 4.4) gives q = 2 or 9 . From figure 4.15 wecan see that q = 2 is the appropriate solution. We ignore the solution q = 9 which occurswhere the upward-sloping branch of the quadratic demand function cuts the supplyfunction for a second time (not shown in figure 4.15). The solution for p is found in theusual way by substituting q = 2 into either the demand or supply function, giving p = 11.(Check this for yourself.)

13

13

1294

QUADRATIC EQUATIONS

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EXAMPLE 4.21

As a second example we suppose that both the demand and supply functions are quadraticin form. The (inverse) functions are

pD = −2q2 − 2q + 35 and

pS = 0.5q2 + q + 3.5

and these are graphed in figure 4.16. Note that in this example the coefficient of q2

in the demand function is negative. This means that the standard U-shape of the quadratic function is inverted, with its turning point a maximum rather than a minimum value as it was in example 4.20. From the graph we can see that this maximum occurs at around q = −0.5, which means that when q is positive, which is thepart of the curve that interests us, the curve is downward sloping, as a demand functionshould be.

A feature of the quadratic supply function in this example is that it becomes steeper as q increases. Thus when q is small, the supply curve is quite flat, so a relatively smallincrease in p induces a relatively large increase in quantity supplied. When q is large, the same size of price increase induces a much smaller increase in quantity supplied. This could be because suppliers are reaching the limits of their productive capacity.

We can find the market equilibrium in the same way as in example 4.20. Given theequilibrium condition pD = pS, we can set the right-hand sides of the demand and supplyfunctions equal to one another, giving

−2q2 − 2q + 35 = 0.5q2 + q + 3.5

This rearranges to

−2.5q2 − 3q + 31.5 = 0

Solving this slightly messy quadratic using the formula (rule 4.4) gives q = 3 or −4.2. There are two solutions because the two curves cut twice, the intersection at q = −4.2 being off the graph to the left in figure 4.16. We discard this negative solution as having no economic meaning. So we are left with q = 3, from which p can be found from either the supply or demand function as p = 11.

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S

pD = –2q2 − 2q + 35

pS = 0.5q2 + q + 3.5

p

q

–10

–5

5

10

15

20

25

30

40

–1 10 2 3 4 5 6

35

whoops!

pS = 2q + 7

pD = 1.5q2 − 15q + 35

–10

0

10

20

30

50

–1 1 2 3 4 5 6 7q

p

40

Figure 4.15 A quadratic inverse demand function with Figure 4.16 Quadratic inverse supply and demand functions.a linear inverse supply function.

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4.14 Economic application 2: costs and revenueA firm’s total cost function

Another application of the quadratic function is to the relationship between a firm’s output andits total costs. At the most general level we can assume that a firm’s total costs, TC, are relatedto its output, q, by some function that we can denote by

TC = f(q) where f( ) denotes some unspecified functional form

What are the most likely shapes for this function? Obviously this is a big subject and here we will attempt only the briefest sketch. An important distinction is between the short run, whenthe firm is assumed to have entered into certain cost commitments such as leasing factory spaceand equipment, and the long run when all such commitments expire and therefore all costs are variable. Considering the short run, a possible shape might be linear: that is, of the form TC = aq + b. An example with a = 1.5 and b = 100 is TC = 1.5q + 100 (see figure 4.17). We assume b, the intercept term, is positive because of our assumption of cost commitments, or fixed costs. The intercept term measures these fixed costs, since they are by definition coststhat have to be met even if output is zero and remain constant whatever the level of output. Weassume a is positive (or possibly zero, in some special cases) because if a were negative thiswould mean that a larger output costs less to produce than a smaller output, which is very hardto imagine.

The problem with the linear short-run total cost function is that it implies that the firm can increase output indefinitely. This seems to conflict with our definition of the short run as a period in which the firm has a given plant size, stock of machinery and so on. A quadratic total cost function, of the form TC = aq2 + bq + c, therefore seems more plausible. An examplecould be

TC = 0.02q2 + 1.5q + 100

(see also figure 4.17). This is the same as the linear TC function in the previous paragraph, but with the addition of a q2 term with a very small coefficient of 0.02. Comparing this with the linear TC function in figure 4.17, we see that the addition of the quadratic term causes the TC curve to turn upwards, increasing steeply as output increases, reflecting the assump-tion that costs rise more rapidly as the firm’s productive capacity is utilized more and more intensively.

A monopolist’s total revenue function

A monopolist is a firm that is the sole source of supply of the product it produces. For reasonsthat we will examine more fully in chapter 8, a firm with a monopoly may find that if it progres-sively lowers its price and therefore sells more and more of its product, total sales revenue risesat first, reaches a maximum value, and then declines as quantity sold increases further. The decline in total revenue beyond a certain point occurs because the revenue loss from the lowerprice begins to outweigh the revenue gain from the increase in quantity sold.

A quadratic function of the form TR = aq2 + bq (where TR = total sales revenue and q = quan-tity sold) is therefore a suitable functional form for modelling this relationship. We assume thatthe parameter a is negative, as is required to make the turning point a maximum rather than aminimum value (see section 4.8 above). We also assume that there is no constant term (c = 0),because if the firm does not sell any output (q = 0), it will not receive any sales revenue (TR = 0).A total revenue function with these properties is

TR = −0.12q2 + 10q

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which is shown in figure 4.18. Total revenue from sales reaches a maximum when q = 42(rounded to the nearest whole unit). Larger quantities can be sold, but only by reducing theprice by so much that total revenue falls. Hopefully this brief discussion has whetted your appetite for further analysis in chapter 8.

Summary of sections 4.13–4.14

The quadratic function may be applied to any area of economic analysis where we have reasonto believe that relationships between economic variables may be non-linear. In section 4.13 weconsidered the market for a good where either the inverse demand function, the inverse supplyfunction, or both, were quadratic functions. In section 4.14 we considered a firm’s short-runtotal cost function which, for sound economic and technological reasons, may be expected tobe non-linear. An inverted quadratic function is also a plausible shape for the total revenuefunction of a monopolist supplier.

The Online Resource Centre shows how supply and demand models and a firm’s total costfunction may be graphed using Excel®. www.oxfordtextbooks.co.uk/orc/renshaw3e/

Progress exercise 4.4

1. Solve the following simultaneous equations, and draw sketch graphs of the functions, indi-cating your solutions.

(a) y = 3x + 6; y = x2 + 2x − 6 (b) y = x2 + 4x − 6; y = −x2 + 2x + 6

2. Given the following supply and demand functions for a good, find the equilibrium price andquantity, and draw sketch graphs of the functions, indicating your solution.

qS = 0.5p2 + p − 8; qD = −2p2 − 2p + 100

3. Given the following inverse supply and demand functions for a good, find the equilibriumprice and quantity, and draw sketch graphs of the functions, indicating your solution.

pS = 10q − 30; pD = −0.5q2 − 8q + 200

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Total revenue from sales reaches a maximumwhen q = 42. Larger quantities can be sold, butonly by reducing the price by so much that totalrevenue falls.

TR = –0.12q2 + 10q

0

50

100

150

200

250

10 20 30 40 50 60 70 80 90 100q

TR

Of the linear and non-linear total cost functions,the non-linear case seems more likely.

TC = 0.02q2 + 1.5q + 100

TC = 1.5q + 100

0

100

200

300

400

500

10 20 30 40 50 60 70 80 90 100q

TC

Figure 4.17 Linear and quadratic total cost functions. Figure 4.18 A quadratic total revenue function.

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QUADRATIC EQUATIONS

Checklist

Be sure to test your understanding of this chapter byattempting the progress exercises (answers are at theend of the book). The Online Resource Centre containsfurther exercises and materials relevant to this chapterwww.oxfordtextbooks.co.uk/orc/renshaw3e/.

In this chapter we have focused on quadratic equations, the simplest form of non-linear rela-tionship. We solved both single and simultaneousquadratic equations and applied these methods to some economic problems. The main topics covered were:

✔ Factorizing quadratic expressions. Understand-ing what quadratic expressions are and how tofactorize them by ‘trial and error’ methods.

✔ Solving quadratic equations. Solving quadraticequations by factorization and also by using theformula. Why some quadratic equations cannot

be factorized and why some have a single, re-peated root.

✔ Quadratic functions and their graphs. Recog-nizing a quadratic function and sketching itsgraph from inspection of its parameter values.

✔ Graphical solution of quadratic equations.Relationship between the algebraic and graph-ical solution of a quadratic equation.

✔ Simultaneous quadratic equations. Solving simultaneous quadratic equations both alge-braically and graphically.

✔ Economic applications. Quadratic functionsapplied to supply and demand models and astotal cost and total revenue functions.

If you have absorbed this material on quadratic equa-tions, you are ready to move on to consider someother types of non-linear function in chapter 5.

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